Contest: Equations as Art 2017

[scottdave]

3987 and 4365 are divisible by 3, therefore their 12th powers are divisible by 3, same for the sum. 4472 is not divisible by 3, and taking the 12th power doesn't change that.

That is the easy method to show it.

 

[Charles Link]

That is the easy method to show it.

A check on simply the last digit does not rule out the possibility that the equality could hold. Before Fermat's last theorem was proven by Andrew Wiles, had someone come up with something like this that worked, it would have been one of the better numerical finds of the century. As I recall, as early as 1970, Fermat's theorem had already been established for exponents $n$ up to 169, so it would have been some very large numbers that would have been necessary to make such a sum.

 

[scottdave]

A check on simply the last digit does not rule out the possibility that the equality could hold. Before Fermat's last theorem was proven by Andrew Wiles, had someone come up with something like this that worked, it would have been one of the better numerical finds of the century. As I recall, as early as 1970, Fermat's theorem had already been established for exponents $n$ up to 169, so it would have been some very large numbers that would have been necessary to make such a sum.

Not the last digit. Sum the digits to see if a multiple of 3.

 

[Charles Link]

Not the last digit. Sum the digits to see if a multiple of 3.

Yes, @scottdave , @mfb 's method is clever.

 

[jfizzix]

$(1+2+3+...+n)^{2}=1^{3} + 2^{3} + 3^{3} +... + n^{3}$

 

[hsdrop]

\begin{matrix}
. & . & . & . & . & . & . & . & . \\
. & P & P & . & . & F & F & F & . \\
. & P & . & P & . & F & .& . & . \\
. & P & P & . & . & F & F & F & . \\
. & P & . & . & . & F & . & . & . \\
. & P & . & . & . & F & . & . & . \\
. & . & . & . & . & . & . & . & . \\
\end{matrix}
I Hope everyone likes it.
I'm also hoping that it falls within the rules as well.

 

[epenguin]

$$\begin{equation} D^{1}\left( uv\right) =uD^{1}v+vD^{1}u \end{equation}$$
$$ \rm {and~ the~ binomial~ expansion~ formula} $$
$$\begin{equation}\left ( a^{1}b^{0}+a^{0}b^{1}\right) ^{n}=\sum ^{n}_{i=0}\binom {n} {i}a^{i}b^{n-i} \end{equation}$$
$$\rm together~ imply~ Leibniz' ~ theorem:$$
$$~D^{n}\left( uv\right) =(D^{0}uD^{1}v+D^{1}uD^0{v})^{n}~ =\sum ^{n}_{i=0}\binom {n} {i}D^nu D^{n-i}v $$
^^ just fits on a page in my preview. It was not necessary for this competition that the equations be true or useful, but whether that is so can be discussed on another thread.
https://www.physicsforums.com/threads/prove-the-leibnitz-rule-of-derivatives.924400/

 

[Demystifier]

$\overbrace{\smile}^{\theta\theta}$

I think this is the only one so far which fulfills the propositions.

 

[martinbn]

What exactly is the definition of equation used for this thread? Some of the things posted, I would call formulae, some expressions and so on.

 

[Demystifier]

At least there must be some wisdom in the symbols to call it equation, like in this one:
$\widehat{\dbinom{\odot_\text{v}\odot}{\wr}}$

 

[Demystifier]

If the above is not counted as an actual equation, then I would mention one of the most difficult unsolved problems in number theory:
$a+b=c$

 

[fresh_42]

If the above is not counted as an actual equation, then I would mention one of the most difficult unsolved problems in number theory:
$a+b=c$

"Unsolved" is debatable. O.k. apparently nobody can really follow the suggested proof, but does this count for "unsolved"?

 

[Greg Bernhardt]

We have a tie between @Orodruin and @MarkFL and someone needs to break it!

 

[Ssnow]

A Srinivasa Ramanujan formula:

$$\frac{1}{1+\frac{e^{-2\pi\sqrt{5}}}{1+\frac{e^{-4\pi\sqrt{5}}}{1+\frac{e^{-6\pi\sqrt{5}}}{1+\cdots}}}}\,=\, \left(\frac{\sqrt{5}}{1+\sqrt[5]{5^{\frac{3}{4}}\left(\frac{\sqrt{5}-1}{2}\right)^{\frac{5}{2}}-1}}-\frac{\sqrt{5}+1}{2}\right)\cdot e^{\frac{2\pi}{\sqrt{5}}}$$

a beautiful combination of $1,2,3,4,5,6$ and other ...
Ssnow

 

[Demystifier]

"Unsolved" is debatable. O.k. apparently nobody can really follow the suggested proof, but does this count for "unsolved"?

I didn't know that there is a suggested proof. Reference?

 

[Baluncore]

a beautiful combination of 1,2,3,4,5,6 and other ...

I see a nest of golden ratios in there, (√5 ± 1 ) / 2.

 

[mfb]

We have a tie between @Orodruin and @MarkFL and someone needs to break it!

12 vs 10 at the moment.

I didn't know that there is a suggested proof. Reference?

The Wikipedia page has a link to it.

 

[scottdave]

We have a tie between @Orodruin and @MarkFL and someone needs to break it!

I vote for @MarkFL s equation.

 

[Charles Link]

Looks like @Ygggdrasil 's takeoff on Fermat's Last theorem has the #3 spot.

 

[Greg Bernhardt]

@Orodruin wins! It is very elegant. Thanks all!